Using the normal distribution, it is found that:
a) The pilot is at the 72th percentile.
b) 19.13% of pilots are unable to fly.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 72.6, \sigma = 2.7[/tex].
Item a:
The percentile is the p-value of Z when X = 74.2, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{74.2 - 72.6}{2.7}[/tex]
Z = 0.59
Z = 0.59 has a p-value of 0.7224.
72th percentile.
Item b:
The proportion that is able to fly is the p-value of Z when X = 78 subtracted by the p-value of Z when X = 70, hence:
X = 78:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{78 - 72.6}{2.7}[/tex]
Z = 2
Z = 2 has a p-value of 0.9772.
X = 70:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70 - 72.6}{2.7}[/tex]
Z = -0.96
Z = -0.96 has a p-value of 0.1685.
0.9772 - 0.1685 = 0.8087 = 80.87%.
Hence the percentage that is unable to fly is:
100 - 80.87 = 19.13%.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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